Question 2: Distribution of Recovery Time from A
Surgery
Time to recovery (in days) after a specific knee surgery procedure.
This follows a typical log-logistic pattern in medical
survival/recovery analysis:
8.23, 12.74, 14.83, 16.61, 18.16, 19.55, 20.80, 21.94, 23.00, 23.98, 24.89, 25.75, 26.56,
27.34, 28.08, 28.79, 29.48, 30.15, 30.81, 31.45, 32.08, 32.70, 33.31, 33.92, 34.53, 35.13,
35.73, 36.33, 36.93, 37.53, 38.14, 38.75, 39.37, 40.00, 40.64, 41.29, 41.95, 42.63, 43.33,
44.05, 44.79, 45.56, 46.36, 47.20, 48.08, 49.02, 50.03, 51.12, 52.32, 53.65
Based on the above data to perform the following analysis.
- Using method of moment estimation to estimate \(\alpha\) and \(\beta\), denoted by \(\hat{\alpha}\) and \(\hat{\beta}\), respectively.
Moment generating function for a log logistic distribution. \[
\mu_k = E[X^k] = \alpha^k B\left(1+\frac{k}{\beta},\,
1-\frac{k}{\beta}\right)
\] The first two moments are \[
\mu_1 = E[X] = \alpha B\left(1+\frac{1}{\beta},\,
1-\frac{1}{\beta}\right)
\] \[
\mu_2 = E[X^2] = \alpha^2 B\left(1+\frac{2}{\beta},\,
1-\frac{2}{\beta}\right)
\] Beta Function Identity \[
B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}
\] Simplifying both moments \[
\mu_1
=
\alpha
\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)
\] \[
\mu_2
=
\alpha^2
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
\] The first two sample moments are \[
m_1 = \frac{1}{n}\sum_{i=1}^n X_i = 34.1922
\] \[
m_2 = \frac{1}{n}\sum_{i=1}^n X_i^2 = 1288.845
\] Substitute theoretical sample moments for moments since
they’re equal \[
34.1922
=
\alpha
\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)
\]
\[
1288.845
=
\alpha^2
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
\] Solve first equation for \(\alpha\) \[
\alpha
=
\frac{34.1922}
{\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)}
\]
Plug \(\alpha\) into second equation
\[
1288.845
=
\left(
\frac{34.1922}
{\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)}
\right)^2
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
\]
Divide both sides by \({m_1^2}\)
(replacing theoretical sample moments with \({m_1}\) and \({m_2}\)) \[
\frac{m_2}{m_1^2}
=
\frac{
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
}{
\left[
\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)
\right]^2
}
\]
Define \[
g(\beta)
=
\frac{
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
}{
\left[
\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)
\right]^2
}
-
\frac{m_2}{m_1^2}
\] Solve \(g(\beta)=0\) to
obtain \(\hat{\beta}\), and then
substitute into \[
\hat{\alpha}
=
\frac{m_1}
{\Gamma\left(1+\frac{1}{\hat{\beta}}\right)
\Gamma\left(1-\frac{1}{\hat{\beta}}\right)}
\]
Code to solve \(g(\beta)=0\) and
obtain both \(\hat{\beta}\) and \(\hat{\alpha}\).
x <- c(8.23, 12.74, 14.83, 16.61, 18.16, 19.55, 20.80, 21.94, 23.00, 23.98, 24.89, 25.75, 26.56,
27.34, 28.08, 28.79, 29.48, 30.15, 30.81, 31.45, 32.08, 32.70, 33.31, 33.92, 34.53, 35.13,
35.73, 36.33, 36.93, 37.53, 38.14, 38.75, 39.37, 40.00, 40.64, 41.29, 41.95, 42.63, 43.33,
44.05, 44.79, 45.56, 46.36, 47.20, 48.08, 49.02, 50.03, 51.12, 52.32, 53.65)
#1st and 2nd moments
m1 = mean(x)
m2 = mean(x^2)
#root finding equation
gb <- function(B){
( (gamma(1 + 2/B) * gamma(1 - 2/B)) / ((gamma(1 + 1/B) * gamma(1 - 1/B))^2) ) - (m2 / m1^2)
}
# since beta > 2
x_vals <- seq(2.01, 10, length.out = 200)
dframe <- data.frame(x = x_vals, y = gb(x_vals))
gk.plot <- ggplot(dframe, aes(x = x, y = y)) +
geom_line(color = "steelblue", size = 1) +
geom_hline(yintercept = 0, linetype = "dashed", alpha = 0.5) + # x-axis
geom_vline(xintercept = 0, linetype = "dashed", alpha = 0.5) + # y-axis
labs(title = "The curve of function g(Beta)",
x = "Beta",
y = "g(Beta)") +
theme(plot.title = element_text(hjust = 0.5),
plot.margin = margin(t = 35, r = 20, b = 30, l = 30, unit = "pt"))
ggplotly(gk.plot)
#find the root (beta value when equation = 0)
B <- uniroot(gb, interval = c(5, 7))$root
#use the beta to find the alpha
alpha <- m1/ (gamma(1+1/B) * gamma(1-1/B))
pander(cbind(beta = B, alpha = alpha))
- Since the moment estimates \(\hat{\alpha}\) and \(\hat{\beta}\) are random, construct
bootstrap sampling distributions for each. To visualize these
distributions, plot separate bootstrap histograms for \(\hat{\alpha}\) and \(\hat{\beta}\). hen, overlay a smooth
density curve on each histogram using Gaussian kernel density
estimation. Finally, describe the patterns of these density curves.
bootstrap <- function (data, statistics, B){
n <- length(data)
stats <- matrix(nrow = B, ncol = 2)
for (i in 1:B){
new_data <- sample(data, n, replace = TRUE)
stats[i,] <- statistics(new_data)
}
return(stats)
}
statistics <- function (data){
m1 = mean(data)
m2 = mean(data^2)
gb <- function(B){
( (gamma(1 + 2/B) * gamma(1 - 2/B)) / ((gamma(1 + 1/B) * gamma(1 - 1/B))^2) ) - (m2 / m1^2)}
x_vals <- seq(2.01, 200, length.out = 100)
dframe <- data.frame(x = x_vals, y = gb(x_vals))
B <- uniroot(gb, interval = c(3, 15))$root
alpha <- m1/ (gamma(1+1/B) * gamma(1-1/B))
c(B, alpha)
}
stats <- bootstrap(x, statistics, 5000)
beta_boot <- stats[, 1]
alpha_boot <- stats[, 2]
hist(beta_boot,
breaks = 20,
probability = TRUE,
xlab = expression(hat(beta)),
main = "Bootstrap Sampling Distribution\n of Beta Hat",
cex.main = 0.9,
col = "lightgray",
border = "white")
lines(density(beta_boot), col = "navy", lwd = 2)

The bootstrap sampling distribution of \(\hat{\beta}\) is roughly bell-shaped with a
little bit of skewness to the right. It’s centered close to the original
MoM estimate, which suggests the estimator is not heavily biased. The
spread of the histogram shows how much \(\hat{\beta}\) would vary if we collected
new samples from the population. Since the distribution isn’t extremely
wide or irregular, it suggests that the MoM estimator for \({\beta}\) is fairly stable for this
sample.
hist(alpha_boot,
breaks = 20,
probability = TRUE,
xlab = expression(hat(alpha)),
main = "Bootstrap Sampling Distribution\n of Alpha Hat",
cex.main = 0.9,
col = "lightgray",
border = "white")
lines(density(alpha_boot), col = "darkred", lwd = 2)

The bootstrap sampling distribution of \(\hat{\alpha}\) is approximately symmetric
and bell-shaped. It’s centered pretty close to the original MoM
estimate, which suggests the estimator is not heavily biased. The spread
of the histogram shows how much \(\hat{\alpha}\) would vary if we collected
new samples from the population. Since the distribution isn’t extremely
wide or irregular, it suggests that the MoM estimator for \({\alpha}\) is fairly stable for this
sample.
---
title: "Assignment 3: Methods of Moment Estimation"
author: "Charlie Morgan"
date: "Due: 2/23/26"
output:
  html_document: 
    toc: yes
    toc_depth: 4
    toc_float: yes
    number_sections: no
    toc_collapsed: yes
    code_folding: hide
    code_download: yes
    smooth_scroll: yes
    theme: lumen
  pdf_document: 
    toc: yes
    toc_depth: 4
    fig_caption: yes
    number_sections: yes
    fig_width: 3
    fig_height: 3
  word_document: 
    toc: yes
    toc_depth: 4
    fig_caption: yes
    keep_md: yes
editor_options: 
  chunk_output_type: inline
---

```{css, echo = FALSE}
#TOC::before {
  content: "Table of Contents";
  font-weight: bold;
  font-size: 1.2em;
  display: block;
  color: navy;
  margin-bottom: 10px;
}


div#TOC li {     /* table of content  */
    list-style:upper-roman;
    background-image:none;
    background-repeat:none;
    background-position:0;
}

h1.title {    /* level 1 header of title  */
  font-size: 22px;
  font-weight: bold;
  color: DarkRed;
  text-align: center;
  font-family: "Gill Sans", sans-serif;
}

h4.author { /* Header 4 - and the author and data headers use this too  */
  font-size: 15px;
  font-weight: bold;
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  color: navy;
  text-align: center;
}

h4.date { /* Header 4 - and the author and data headers use this too  */
  font-size: 18px;
  font-weight: bold;
  font-family: "Gill Sans", sans-serif;
  color: DarkBlue;
  text-align: center;
}

h1 { /* Header 1 - and the author and data headers use this too  */
    font-size: 20px;
    font-weight: bold;
    font-family: "Times New Roman", Times, serif;
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}

h2 { /* Header 2 - and the author and data headers use this too  */
    font-size: 18px;
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}

h3 { /* Header 3 - and the author and data headers use this too  */
    font-size: 16px;
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    color: navy;
    text-align: left;
}

h4 { /* Header 4 - and the author and data headers use this too  */
    font-size: 14px;
  font-weight: bold;
    font-family: "Times New Roman", Times, serif;
    color: darkred;
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}

/* Add dots after numbered headers */
.header-section-number::after {
  content: ".";

body { background-color:white; }

.highlightme { background-color:yellow; }

p { background-color:white; }

}
```

```{r setup, include=FALSE}
# code chunk specifies whether the R code, warnings, and output 
# will be included in the output files.
if (!require("knitr")) {
   install.packages("knitr")
   library(knitr)
}
if (!require("pander")) {
   install.packages("pander")
   library(pander)
}
if (!require("ggplot2")) {
  install.packages("ggplot2")
  library(ggplot2)
}
if (!require("tidyverse")) {
  install.packages("tidyverse")
  library(tidyverse)
}

if (!require("plotly")) {
  install.packages("plotly")
  library(plotly)
}
####
knitr::opts_chunk$set(echo = TRUE,       # include code chunk in the output file
                      warning = FALSE,   # sometimes, you code may produce warning messages,
                                         # you can choose to include the warning messages in
                                         # the output file. 
                      results = TRUE,    # you can also decide whether to include the output
                                         # in the output file.
                      message = FALSE,
                      comment = NA
                      )  
```
 
 \
 
## **Assignment Objectives** 

* Master the fundamental concepts of point estimation and performance metrics

* Understand the theoretical foundation of the method of moments estimator (MME)

* Implement MME in R, incorporating numerical approximation methods

\

**Log-logistic Distribution**

The log-logistic distribution (also known as the Fisk distribution) is a continuous probability distribution that is particularly useful in contexts where data exhibit non-negative, skewed behavior and where the hazard rate is unimodal (increases to a peak and then decreases). It has been widely used in the areas such as survival analysis and reliability engineering, environmental science, economics, pharmacology, finance and risk management, etc. 

For given shape parameter $\beta$ and scale parameter $\alpha$, the cumulative distribution function

$$
F(x) = \frac{1}{1+(x/\alpha)^{-\beta}}
$$

As an exercise, you can derive the density in the following form

$$
f(x) = \frac{(\beta/\alpha)(x/\alpha)^{\beta-1}}{[1+(x/\alpha)^\beta]^2}, \ \ \text{ for } \ \ x > 0.
$$

After some algebra, we can find the $k$th moment

$$
\mu_k = E[X^k] = \alpha^k B\left(1+\frac{k}{\beta}, 1 - \frac{k}{\beta} \right).
$$

This assignment will focus on finding MME of parameters $\alpha$ and $\beta$ based on a real-world application data set.


\

## **Question 1: Derive the log-logistic density function **

Given the CDF of the two-parameter log-logistic distribution

$$
F(x) = \frac{1}{1+(x/\alpha)^{-\beta}}.
$$
$$
\begin{aligned}
F(x)
&= \frac{1}{1 + (x/\alpha)^{-\beta}} \\[6pt]
&= \left(1 + (x/\alpha)^{-\beta}\right)^{-1} \\[10pt]

f(x)
&= \frac{d}{dx}F(x) \\[6pt]
&= -\left(1 + (x/\alpha)^{-\beta}\right)^{-2}
    \cdot
    \frac{d}{dx}\left(1 + (x/\alpha)^{-\beta}\right) \\[10pt]

&= -\left(1 + (x/\alpha)^{-\beta}\right)^{-2}
    \cdot
    \left(0 + (-\beta)(x/\alpha)^{-\beta-1}\cdot \frac{1}{\alpha}\right) \\[10pt]

&= \frac{\beta (x/\alpha)^{-(\beta+1)}}
        {\alpha \left(1 + (x/\alpha)^{-\beta}\right)^2} \\[10pt]

&= \frac{\beta (x/\alpha)^{\beta-1}}
        {\alpha \left(1 + (x/\alpha)^{\beta}\right)^2} \\[12pt]
&= \frac{(\beta/\alpha)(x/\alpha)^{\beta-1}}{[1+(x/\alpha)^\beta]^2}, \ \ \text{ for } \ \ x > 0.
\end{aligned}
$$
\

## **Question 2: Distribution of Recovery Time from A Surgery**

Time to recovery (in days) after a specific knee surgery procedure. This follows a typical **log-logistic pattern** in medical survival/recovery analysis:

```
8.23, 12.74, 14.83, 16.61, 18.16, 19.55, 20.80, 21.94, 23.00, 23.98, 24.89, 25.75, 26.56, 
27.34, 28.08, 28.79, 29.48, 30.15, 30.81, 31.45, 32.08, 32.70, 33.31, 33.92, 34.53, 35.13, 
35.73, 36.33, 36.93, 37.53, 38.14, 38.75, 39.37, 40.00, 40.64, 41.29, 41.95, 42.63, 43.33, 
44.05, 44.79, 45.56, 46.36, 47.20, 48.08, 49.02, 50.03, 51.12, 52.32, 53.65
```
Based on the above data to perform the following analysis.

a) Using method of moment estimation to estimate $\alpha$ and $\beta$, denoted by $\hat{\alpha}$ and $\hat{\beta}$, respectively.

Moment generating function for a log logistic distribution.
$$
\mu_k = E[X^k] = \alpha^k B\left(1+\frac{k}{\beta},\, 1-\frac{k}{\beta}\right)
$$
The first two moments are
$$
\mu_1 = E[X] = \alpha B\left(1+\frac{1}{\beta},\, 1-\frac{1}{\beta}\right)
$$
$$
\mu_2 = E[X^2] = \alpha^2 B\left(1+\frac{2}{\beta},\, 1-\frac{2}{\beta}\right)
$$
Beta Function Identity
$$
B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}
$$
Simplifying both moments
$$
\mu_1
=
\alpha
\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)
$$
$$
\mu_2
=
\alpha^2
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
$$
The first two sample moments are
$$
m_1 = \frac{1}{n}\sum_{i=1}^n X_i = 34.1922
$$
$$
m_2 = \frac{1}{n}\sum_{i=1}^n X_i^2 = 1288.845
$$
Substitute theoretical sample moments for moments since they're equal
$$
34.1922
=
\alpha
\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)
$$

$$
1288.845
=
\alpha^2
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
$$
Solve first equation for $\alpha$
$$
\alpha
=
\frac{34.1922}
{\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)}
$$

Plug $\alpha$ into second equation
$$
1288.845
=
\left(
\frac{34.1922}
{\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)}
\right)^2
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
$$

Divide both sides by ${m_1^2}$ (replacing theoretical sample moments with ${m_1}$ and ${m_2}$)
$$
\frac{m_2}{m_1^2}
=
\frac{
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
}{
\left[
\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)
\right]^2
}
$$

Define
$$
g(\beta)
=
\frac{
\Gamma\left(1+\frac{2}{\beta}\right)
\Gamma\left(1-\frac{2}{\beta}\right)
}{
\left[
\Gamma\left(1+\frac{1}{\beta}\right)
\Gamma\left(1-\frac{1}{\beta}\right)
\right]^2
}
-
\frac{m_2}{m_1^2}
$$
Solve $g(\beta)=0$ to obtain $\hat{\beta}$, and then substitute into
$$
\hat{\alpha}
=
\frac{m_1}
{\Gamma\left(1+\frac{1}{\hat{\beta}}\right)
\Gamma\left(1-\frac{1}{\hat{\beta}}\right)}
$$

Code to solve $g(\beta)=0$ and obtain both $\hat{\beta}$ and $\hat{\alpha}$.
```{r}
x <- c(8.23, 12.74, 14.83, 16.61, 18.16, 19.55, 20.80, 21.94, 23.00, 23.98, 24.89, 25.75, 26.56, 
27.34, 28.08, 28.79, 29.48, 30.15, 30.81, 31.45, 32.08, 32.70, 33.31, 33.92, 34.53, 35.13, 
35.73, 36.33, 36.93, 37.53, 38.14, 38.75, 39.37, 40.00, 40.64, 41.29, 41.95, 42.63, 43.33, 
44.05, 44.79, 45.56, 46.36, 47.20, 48.08, 49.02, 50.03, 51.12, 52.32, 53.65)

#1st and 2nd moments
m1 = mean(x)
m2 = mean(x^2)

#root finding equation
gb <- function(B){
  ( (gamma(1 + 2/B) * gamma(1 - 2/B)) / ((gamma(1 + 1/B) * gamma(1 - 1/B))^2) ) - (m2 / m1^2)
}

# since beta > 2
x_vals <- seq(2.01, 10, length.out = 200)
dframe <- data.frame(x = x_vals, y = gb(x_vals))

gk.plot <- ggplot(dframe, aes(x = x, y = y)) +
  geom_line(color = "steelblue", size = 1) +
  geom_hline(yintercept = 0, linetype = "dashed", alpha = 0.5) +  # x-axis
  geom_vline(xintercept = 0, linetype = "dashed", alpha = 0.5) +  # y-axis
  labs(title = "The curve of function g(Beta)",
       x = "Beta", 
       y = "g(Beta)") +
    theme(plot.title = element_text(hjust = 0.5),
        plot.margin = margin(t = 35, r = 20, b = 30, l = 30, unit = "pt"))
ggplotly(gk.plot)
```

```{r}
#find the root (beta value when equation = 0)
B <- uniroot(gb, interval = c(5, 7))$root

#use the beta to find the alpha
alpha <- m1/ (gamma(1+1/B) * gamma(1-1/B))
pander(cbind(beta = B, alpha = alpha))
```

b) Since the moment estimates $\hat{\alpha}$ and $\hat{\beta}$ are random, construct bootstrap sampling distributions for each. To visualize these distributions, plot separate bootstrap histograms for $\hat{\alpha}$ and $\hat{\beta}$.  hen, overlay a smooth density curve on each histogram using Gaussian kernel density estimation. Finally, describe the patterns of these density curves.

```{r}
bootstrap <- function (data, statistics, B){
  n <- length(data)
  stats <- matrix(nrow = B, ncol = 2)
  for (i in 1:B){
    new_data <- sample(data, n, replace = TRUE)
    stats[i,] <- statistics(new_data)
  }
  return(stats)
}
  
statistics <- function (data){
  m1 = mean(data)
  m2 = mean(data^2)
  
  gb <- function(B){
  ( (gamma(1 + 2/B) * gamma(1 - 2/B)) / ((gamma(1 + 1/B) * gamma(1 - 1/B))^2) ) - (m2 / m1^2)}
  
  x_vals <- seq(2.01, 200, length.out = 100)
  dframe <- data.frame(x = x_vals, y = gb(x_vals))
  
  B <- uniroot(gb, interval = c(3, 15))$root
  alpha <- m1/ (gamma(1+1/B) * gamma(1-1/B))
  
  c(B, alpha)
}

stats <- bootstrap(x, statistics, 5000)

beta_boot  <- stats[, 1]
alpha_boot <- stats[, 2]

hist(beta_boot,
     breaks = 20,
     probability = TRUE,
     xlab = expression(hat(beta)),
     main = "Bootstrap Sampling Distribution\n of Beta Hat",
     cex.main = 0.9,
     col = "lightgray",
     border = "white")

lines(density(beta_boot), col = "navy", lwd = 2)
```

The bootstrap sampling distribution of $\hat{\beta}$ is roughly bell-shaped with a little bit of skewness to the right. It’s centered close to the original MoM estimate, which suggests the estimator is not heavily biased. The spread of the histogram shows how much $\hat{\beta}$ would vary if we collected new samples from the population. Since the distribution isn’t extremely wide or irregular, it suggests that the MoM estimator for ${\beta}$ is fairly stable for this sample.
```{r}
hist(alpha_boot,
     breaks = 20,
     probability = TRUE,
     xlab = expression(hat(alpha)),
     main = "Bootstrap Sampling Distribution\n of Alpha Hat",
     cex.main = 0.9,
     col = "lightgray",
     border = "white")

lines(density(alpha_boot), col = "darkred", lwd = 2)
```

The bootstrap sampling distribution of $\hat{\alpha}$ is approximately symmetric and bell-shaped. It’s centered pretty close to the original MoM estimate, which suggests the estimator is not heavily biased. The spread of the histogram shows how much $\hat{\alpha}$ would vary if we collected new samples from the population. Since the distribution isn’t extremely wide or irregular, it suggests that the MoM estimator for ${\alpha}$ is fairly stable for this sample.
